Spectral decomposition and invertibility

Question: Let $\displaystyle T$ be a normal operator on a finite-dimensional complex inner product space $\displaystyle V$. Consider spectral decomposition $\displaystyle T = \lambda_1 T_1 + ... + \lambda_k T_k$ and prove that $\displaystyle T$ is invertible if and only if $\displaystyle \lambda_i \neq 0$ for $\displaystyle i \leq i \leq k$.

Any pointers would be helpful - I am unable to begin currently.

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Originally Posted by Last_Singularity

Question: Let $\displaystyle T$ be a normal operator on a finite-dimensional complex inner product space $\displaystyle V$. Consider spectral decomposition $\displaystyle T = \lambda_1 T_1 + ... + \lambda_k T_k$ and prove that $\displaystyle T$ is invertible if and only if $\displaystyle \lambda_i \neq 0$ for $\displaystyle i \leq i \leq k$.

Any pointers would be helpful - I am unable to begin currently.

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Well, this is ALWAYS true: any operator (of a finite-dimensional vector space $\displaystyle V$) over any field is invertible iff all its eigenvalues are different from zero, and the proof is painfully simple: $\displaystyle T$ is NOT invertible iff $\displaystyle Ker(T)\ne {0}\,\Longleftrightarrow\,\exists\,0\ne v\in V\,\,\,s.t.\,\,Tv=0=0\cdot v\,\Longleftrightarrow\,0$ is an eigenvaue of $\displaystyle T$.

Tonio